(A) Since the battery remains connected,the potential difference across the plates remains constant. Therefore,$V = V_0$.The electric field is given b

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(A) Since the battery remains connected,the potential difference across the plates remains constant. Therefore,$V = V_0$.The electric field is given by $E = V/d$. Since $V$ and the separation $d$ remain unchanged,$E = E_0$.Let the initial capacitance be $C_0$. When a dielectric slab of constant $K$ is inserted,the new capacitance becomes $C = K C_0$,where $K > 1$.The charge on the capacitor is $Q = CV = (K C_0) V_0 = K Q_0$. Since $K > 1$,it follows that $Q > Q_0$.The energy stored in the capacitor is $U = (1/2) CV^2 = (1/2) (K C_0) V_0^2 = K U_0$. Since $K > 1$,it follows that $U > U_0$.Thus,the correct relation is $Q > Q_0$.

Class 12 Physics Electric Potential and Capacitance Effect of Dielectric Inside Capacitor

Medium

$A$ parallel plate capacitor is connected to a battery. The quantities charge,voltage,electric field,and energy associated with the capacitor are given by $Q_0, V_0, E_0$,and $U_0$ respectively. $A$ dielectric slab is introduced between the plates of the capacitor,but the battery remains connected. The corresponding quantities now given by $Q, V, E$,and $U$ related to the previous ones are:

A
$Q > Q_0$
B
$V > V_0$
C
$E > E_0$
D
$U < U_0$

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

For changing the capacitance of a given parallel plate capacitor,a dielectric material of dielectric constant $K$ is used,which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} d$,where $d$ is the separation between the plates of the parallel plate capacitor. The new capacitance $(C')$ in terms of original capacitance $(C_{0})$ is given by the following relation:

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Half of the space between the plates of a parallel-plate capacitor is filled with a dielectric material of dielectric constant $K$. The remaining half contains air. The capacitor is now given a charge $Q$. Then:

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Four metallic plates $A, B, C$ and $D$ of the same size with the same separation between them are arranged as shown in the figure. Dielectric slabs of dielectric constant $K = 2$ are placed between $B, C$ and $C, D$ respectively. Plates $B$ and $D$ are connected together. The effective capacitance between $A$ and $C$ is (Assume capacitance of each pair of plates without dielectric is $C$):

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The distance between the plates of a parallel plate capacitor is $8\,mm$ and the potential difference $(P.D.)$ is $120\,V$. If a $6\,mm$ thick slab of dielectric constant $K = 6$ is introduced between its plates,then:

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$A$ slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$,where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

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Half of the space between the plates of a parallel-plate capacitor is filled with a dielectric material of dielectric constant $K$. The remaining half contains air. The capacitor is now given a charge $Q$. Then:

The distance between the plates of a parallel plate capacitor is $8\,mm$ and the potential difference $(P.D.)$ is $120\,V$. If a $6\,mm$ thick slab of dielectric constant $K = 6$ is introduced between its plates,then:

$A$ slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$,where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

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